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Put $\delta =\exp (-{\epsilon }^{-C})$, with the absolute constant $C$ chosen sufficiently large below, and suppose for a contradiction that no $g_2$ with $0\le g_2\le 1$ satisfies $\Vert g_1-g_2{\Vert }_{\mathcal{F}}\le \epsilon$.

LetThe set $K$ is compact and convex, while $B$ is closed and convex, so $K+B$ is closed and convex. We use the following finite-dimensional form of the Hahn-Banach separation theorem: if a point lies outside a nonempty closed convex set, there is a linear functional whose value at the point is strictly greater than its supremum over that set. Identifying linear functionals on ${\mathcal{H}}_N$ through the inner product, there is therefore a function $\psi$ such thatThe separating functional cannot have zero dual norm, so rescale it to make $\Vert \psi {\Vert }_{\mathcal{F}}^{\ast }=1$. Because $\mathcal{F}$ is closed, convex, and symmetric, the finite-dimensional Bipolar theorem for a dual pair says that the unit ball of $\Vert \cdot {\Vert }_{\mathcal{F}}^{\ast }$ is precisely $\mathcal{F}$. Thus $\psi \in \mathcal{F}$ and $|\psi (n)|\le 1$.

The support function of $K$ is obtained by choosing $g(n)=1$ where $\psi (n)>0$ and $g(n)=0$ where $\psi (n)<0$. In terms of the positive part of a real-valued function ${\psi }_{+}=\max (0,\psi )$, the separating inequality becomesSince $0\le g_1\le \nu$, pointwise we have $g_1\psi \le \nu {\psi }_{+}$, and hence

Apply the supplied polynomial approximation of the positive part to $\psi$. If $P_{\epsilon }(t)={\sum }_{i=0}^m{a}_i{t}^i$, then its uniform approximation error and $\mathbb{E}\nu \le 1$ implyThe constant function $1$ and $\psi$ belong to the dual unit ball. By the assumed submultiplicativity, $\Vert {\psi }^i{\Vert }_{\mathcal{F}}^{\ast }\le 1$ for every $i\ge 0$. Dual seminorm therefore givesThe stated coefficient bound, with $C$ in $\delta$ chosen larger than the absolute constant in that bound, makes this last quantity at most $\epsilon /50$. Together with the polynomial-approximation error, this contradicts $⟨\nu -1,{\psi }_{+}⟩>\epsilon$. The required $g_2$ consequently exists. This proves the dense model theorem for a multiplicative test family.

The function $\varphi (x)={\max }_{v\in C}⟨x,v⟩$ is the support function ${\sigma }_C$. For a nonempty compact convex set,so its convex conjugate is the indicator function ${\iota }_C$. Applying the Moreau decomposition,Multiplication of an indicator function by a positive scalar does not change it, and its proximal operator is the Euclidean projection onto a convex set. Therefore

Take $a=1$ and $b=k$, sois the capped simplex. A linear objective over this convex polytope attains its maximum at a zero-one extreme point. Choosing the $k$ coordinates at which $x_i$ is largest givesEquivalently, an exchange of weight from a smaller component to a larger one never decreases the objective. Thus the sum of the largest components is the support function ${\sigma }_C$.

Part c now givesBy the projection onto a box-constrained hyperplane, $v=P_C(y/t)$ hasConsequently the proximal operator is evaluated by solving this one-dimensional equation for $\nu$, then substituting the resulting projection.

The sum of the $k$ largest components of $x\in {\mathbb{R}}^n$ is the support function of the capped simplex: ${\sum }_{i=1}^k{x}_{[i]}=\max \{x^{T}v:0\le v\le 1,{\sum }_i{v}_i=k\}$.